Chapter 16.5 - Right angled stairs

In the previous section we discussed the special precautions to be taken in the arrangement of reinforcements at Landings. In this section we continue the discussion on stairs in general. We will now discuss about a different type of stair.

Right angled stairs
The stair shown in fig.16.30 below has two flights. The second flight is deviating from the first flight at an angle of 90^o.

Fig.16.30
Stair taking a right angled turn

The first flight AB is supported on the ground at A, and on the wall-2 at B. The second flight CD is supported on wall-1 at C and on a beam at D. [At the time of construction, the walls 1 and 2 are stopped at the level of the bottom surface of the intermediate landing, so that the landing can be extended into the walls. After the casting and curing of landing and flight AB, the wall construction is resumed. The bars for CD are extended from the landing because CD can be casted only after completing wall-2, as it rests on the beam at D. The safety and stability of the beam at D should be carefully checked.] The bearing of the landing on the two walls can be seen in the plan view given below:

Fig.16.31
Plan view of right angled stair

For analysis, we have to draw the line diagram of each of the flights separately. The line diagrams are drawn from support to support. That is., from A to B, and from C to D. This is shown in figs.16.32 and 16.33 below.

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We have to make a modification to the loads on the intermediate landing. It can be explained as follows: • We are designing the flights separately. 
• When we assign loads to each flights separately, the loads coming on the intermediate landing will be assigned to both the flights.
• If we assign the full landing load w₂ to one flight, it will mean that, the full load w₂ acting on the landing is resisted by that flight only. It will also mean that the other flight does not have to resist w₂
• But we know that both flights will take part in resisting w₂. In other words, w₂ contributes towards the bending of both the flights.
• So if we assign the full w₂ to any one stair, there will be under estimation of the load on the other stair
• On the other hand, if we assign full w₂ to both the flights, there will be over estimation. This is because, neither of the flights have to resist the full w₂.
• So we give a part of w₂ for one flight, and the remaining part for the other flight. In usual design practice, this 'part' is exact 50%. That is., 0.5w₂ is given to one flight and 0.5w₂ is given to the other flight.
• Thus, we first calculate w₂ on the landing as usual, and then assign 0.5w₂ for the landing common to each flight. This is shown in the line diagrams below.

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Fig.16.32Line diagram for flight AB

Fig.16.33
Line diagram for flight CD

It may be noted that the modification factor of 0.5 is not applied to w2 on the topmost landing. This is because, it is not a common landing, and so, duplication does not occur. Also note that, the last step of flight AB should be of reinforced concrete as shown in fig.16.32. This is to ensure the smooth continuation of the main bars of flight CD into wall-1.
Once the line diagrams are prepared, we can do the analysis to determine the maximum bending moments, and then do the design as usual.

Stairs supported on landings
In the cases that we saw so far, the flights were supported on walls or beams. Now we will consider a type of longitudinal stair in which the the flights are supported by 'landing slabs'. To fully understand such a support, we have to get more details of the building. So let us start from the 'Key plan'. Fig.16.34 below shows a part of the key plan of a two storey building. The plan of the stairs in Floor 2 is shown within the key plan.

Fig.16.34
Portion of the key plan showing the Staircase portion

The sectional view is shown in the fig.16.35 below:

Fig.16.35
Sectional elevation

The beams and columns in the above figs. are named according to the specifications given in SP34. For example, COL 2 Q1 indicates the column Q1 (the column at the North East corner of the building) in Floor 2.

We can think about the supports of the stair in this way:
Flight AB can be supported at A on Bm 4 (of Floor1). At B, it can be supported on a brick wall built above Bm 3 (parallel to the Risers). Flight CD can be supported at C on this brick wall. At D, it can be supported on Bm 4 (of Floor 2) So the flights will be spanning between these beams and the brick wall on Bm 3. That is.,
• Flight AB spans between Bm4 and wall above Bm3 [Both Bm4 and Bm3 are at level 1]
• Flight CD spans between wall above Bm3 and Bm4 [Bm3 is at level 1 and Bm4 is at level 2]
But such a brick wall above Bm3 is not shown in the sectional view. Also, at A and D, the the flights are not shown to have any bearing into the beams Bm4 on either levels. So how are the flights being supported? For the answer, we will look at the views given in the next section.

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Chapter 16.4 - Bars for landings in stairs

In the previous section we designed a stair and saw it's reinforcement details. In this section we will discuss more details about the arrangement of bars.

In the fig.16.22 which shows flight AB, the bottom layer bar (bar type 'a') in the sloping portion becomes the top layer in the landing portion. For the bottom layer of landing, extra bars are given. Similar arrangement can be seen in flight CD (fig.16.23) also. We will now discuss the reason for giving such an arrangement. Consider fig. 16.24 below:

Fig.16.24
Stair bars without embedment

In the fig.16.24, the bottom bar from the sloping portion continues into the landing in such a way that it is the bottom bar in the landing also. When the loads are applied on the slab, the bar will be in tension, and it will try to straighten up. Only the concrete cover is present there to resist this tendency of the bar to straighten up. This concrete cover does not have enough thickness to adequately resist this tendency, and cracks may develop. So we must extend this bar to embed it into a 'mass of concrete'. To achieve this embedment, the bar is taken up to near the top surface of the landing, and then a bend is given to make it horizontal. The measurements required for this embedment is shown in the fig.16.25 below:

Fig.16.25
Length of embedment required by the bars

We can see that, when the type 'a' bars become the top bars of the landing, the bottom portion of the landing is left with out any bars. So some extra bars (denoted as bar type 'b') are given at the bottom layer of the landing. These bars should have the same diameter and spacing as 'a' type bars. The 'b' bars also should have sufficient embedment. So they are taken upto near the top surface of the sloping slab, and then given a bend, to make them parallel to the slope. 

The point of intersection of 'a' and 'b' is taken as the 'critical' point. The specified embedment should be measured from this point. In the fig., the length required is specified as 'Ld(min)'. Why is it specially mentioned as 'min'? The explanation is as follows: The bars we are considering are 'top bars'. Their main purpose is to resist the hogging moment (that can possibly arise even at a simple support due to partial fixity). So they must have the specified length which is more than the length over which the hogging moment can possibly act. Thus we have two lengths to consider:
• The length required for resisting the hogging moment. Which is taken as 0.25l for general cases
• The length required for the embedment to prevent straightening up. Which is Ld
The largest of the above two lengths should be used. This will satisfy both the requirements. So the mention of 'min' tells us to take both criteria into consideration. It may be noted that in Limit state design, Ld is the unique value that we saw in a previous chapter. Also, Ld  should be provided on both sides of the critical point.

Now we consider the landing portion at ‘C’ (the intermediate landing) for the flight CD in fig.16.23. Here the bar type ‘a’ will not try to straighten up. So it does not require any extra embedment. So these bars continue as bottom bars into the landing. But at the support at this intermediate landing, hogging moments can occur if a wall is constructed above the landing (causing partial fixity), as shown in the fig. below:

Fig.16.25 (a)
Hogging moment at support

We can give top bars in the landing which will continue as top bar in the sloping portion also. But when the hogging moment occurs, these bars will be in tension, and will try to straighten up. So we must give two sets of bars (types ‘c’ and ‘d’) as shown in fig.16.25(a) above.
It must be noted that in fig.16.23, the different sets of bars are shown in separate layers only for clarity. In the actual structure, they will be in same layer as shown in the fig. below:

Fig.16.25(b)
Types of bars in same layers

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Stairs with overhanging Landings
Now, we can discuss the arrangement of bars in another type of longitudinal stairs. In this type, the supports are at the ends of the sloping slab as shown in the fig.16.26 below:

Fig.16.26
Supports at the ends of sloping slab

From the fig., we can see that the supports are at the ends of the sloping slabs. The intermediate landing and the top landing are overhanging beyond the supports. In other words, the landings are cantilevers. This type of stairs are more economical. Let us see how this economy is achieved: As shown earlier, the thickness of the waist slab is taken as 1/20 of the effective span. In the fig. above, the effective span is reduced because, the supports are now nearer to each other. So the waist slab thickness can be reduced. The bending moment will also be reduced because of the reduced effective span.

A cantilever structure will produce more bending moment than a simply supported structure. So at a glance, we may feel that more steel will be required for resisting the bending moment from the cantilevers. But here, the cantilevering span is small when compared to the simply supported span between the supports. Also the load on the cantilever landings is less than the load on the sloping portion.
However, it is important to note that hogging moments will be produced at the supports because of the cantilever action. This is shown in the figs. below:

Fig.16.27
Bending moment diagram for flight AB

Fig.16.27
Bending moment diagram for flight CD

From the above figs., it is clear that hogging moments will be present at the supports in this type of stairs. The values of maximum hogging moments and sagging moments can be easily calculated from basic principles. Then we can determine the steel required to resist these moments. The steel required to resist the hogging moments should be given as top steel at the supports. The arrangement of bars for this type of stairs is shown in the figs. below:

Fig.16.28
Arrangement of bars for flight AB

Fig.16.29
Arrangement of bars for flight CD

In the above figs., a new type of bar denoted as 'd' is given as top bars at all supports where overhang is present. These are the bars which resist the hogging moment. They must have sufficient embedment as indicated by 'y' in the figs. Also note that these bars are taken to the farther face of the sloping slab, and then bent to make them parallel to the slope. This is to give maximum embedment inside concrete.

In the next section, we will discuss about another type of longitudinal stairs in which, the second flight takes a right angled turn from the first flight.

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Chapter 16.3 - Reinforcement details of staircase

In the previous section we saw the self wt. of stair slab and the steps. Now we will see the other loads coming on the stairs.

Self weight of finishes (16.6)
The self weight of finishes applied over the steps should be considered in the design. It can be obtained from data books or relevant codes. Usually it varies from 0.5 to 1.0 kN/m². This load obtained from data books or codes, is assumed to act vertically on a horizontal plane. So there is no need to make any modifications, and we can apply it directly.

Live Loads (16.7)
The Live loads acting on the stairs can also be obtained from the data books or relevant codes. IS 875: 1987(part II) recommends a uniformly distributed load of 5 kN/m². This load is to be applied on both the sloping portion and the horizontal landing. In buildings such as residences, where the specified Live loads on the floors do not exceed 2 kN/m², and the stairs are not liable to be overcrowded, the Live load can be taken as 3kN/m². As in the case of self wt. of finishes, the LL obtained from data books or codes, is assumed to act vertically on a horizontal plane, and so there is no need make any modifications, and we can apply it directly.

Loads on Landing
The above items 16.3, 16.5, 16.6 and 16.7 are the four loads that come on the sloping slab (Going) of a flight. Let us now see the loads that come on the landing.

The landing is horizontal, and so there is no need to make any modifications for slope. The volume of a 1 x1m square block of landing slab is 1 x 1 x t = t m³. Multiplying this by the weight density of reinforced concrete will give the weight. So we get 25t kN/m². (16.8)
• The loads coming from the finishes can be taken as the same value (16.6) that we saw for the sloping portion.
• The LL can also be taken as the same load (16.7) for sloping portion.

So we have seen all the loads coming on the stair. The loads that we determine using the above methods are characteristic loads. They must be multiplied by the appropriate load factors to obtain the factored loads.
All these loads are calculated for a 1 x1m square area. But we are considering a beam (strip of slab) of width 1m, as shown in fig.16.16. So the above load will also be the load acting on every 1m length of the beam. In other words, the load on a 1 x1 m sq. area is also the UDL acting per meter length of the beam. So we can represent our beam as shown in the fig. below:

Fig.16.20
Loads acting on the stair

In the above fig.,
Eq.16.9: w₁ (Load per unit length on Sloping portion) = Load factor x {sum of values obtained from (16.3, 16.5, 16.6 and 16.7)}
Eq.16.10: w₂ (Load per unit length on Landing)= Load factor x {sum of values obtained from (16.8, 16.6 and 16.7)}
Note that in Eq.16.10,
• The loads from finishes (16.6) and the LL (16.7) remains the same as that of the sloping portion.
• 16.5 is absent because there are no steps in the landing.
• 16.3 (wt. of inclined slab) is replaced by 16.8 (wt. of horizontal slab)
So w₂ will have a lesser value than w₁ because 16.8 will always be lesser than 16.3.

Thus we can calculate the loads and draw the diagram shown in fig.16.20 with all the details. Based on that fig., we can draw the BM and Shear force diagrams.

In fig.16.14 of the previous section, we have considered a strip from support to support. The fig. is shown again for convenience.

Fig.16.14
1 m wide strip

One support is a masonry wall. The other is the end of the slab which is given an increased thickness. It rests on the foundation. The strip bends between these two supports. The risers (not shown in the fig.) of this stair are all 'parallel to the two supports'. Such stairs which bend between 'supports which are parallel to the risers' are called Longitudinal stairs. The other type is the stairs which bend between 'supports which are perpendicular to the risers' are called Transverse stairs. These two types are shown schematically in the figs. below:

Schematic diagram for Longitudinal stairs

Schematic diagram for Transverse stairs

We will discuss about transverse stairs in later sections.

So we have a longitudinal stairs with two flights AB and CD. Flight AB has one Going and one Landing. Many times in practice, we will come across this type with one Going and one landing, which can be represented by the line diagram shown in fig.16.20 above. So it will be convenient if we derive the general equations for BM and shear forces. These equations are given below:
Reaction at support A is given by:

Eq.16.11

Reaction at support B is given by:
Eq.16.12

The bending moment at any point at a distance x from support A is given by:
Eq.16.13

If we differentiate this equation, we will get the equation for the shear force at any point at a distance x from the support. This is given below:

Eq.16.14

Maximum bending moment occurs at a place where the shear force is equal to zero. So we equate the above equation 16.14 to zero and solve for x. Then we put this value of x in 16.13 and find the value of maximum BM. The reinforcement is then designed for this maximum BM.

We have just seen the procedure for the analysis and design of a flight with one Going and one Landing. Now we will see the procedure for a flight with one Going and two landings. The method of calculation of loads are the same. When the loads are calculated, we can draw the line diagram as shown below:

Fig.16.21
Loads acting on a flight with one Going and two landings

The calculations of BM and SF are easy in this case because the beam and the loadings are symmetrical. (Note that, the lengths of both the landings are taken as l₂. If they are not equal, there will not be symmetry). As before, we will write the equations as given below:
Reactions at supports C and D is given by

Eq.16.15

As the beam and the loadings are symmetrical, the maximum BM will be at the mid span. So there is no need to find the point where the SF is equal to zero. The maximum BM is given by:
Eq.16.16

Where l = l₁ + 2l₂

When the maximum bending moment is calculated, we can design the reinforcement required to resist that bending moment. So we will now do the detailed analysis and design of our stair. It is given as:

Solved example 16.1
The arrangement of bars according to the design in the above solved example is shown in the figs.16.22 and 16.23 given below:

Fig.16.22
Reinforcement details of flight AB

Fig.16.23
Reinforcement details of flight CD

For flight AB, the main bars of 12 mm dia. are provided at 250 mm c/c. For the flight CD, the main bars of 12 mm dia. are provided at 150 mm c/c. In the next section, we will discuss some details about the above two figs.

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